Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 1 - Section 1.7 - Absolute Value Equations and Inequalities - 1.7 Exercises - Page 118: 17


$x=\left\{ \dfrac{7}{3},3 \right\}$

Work Step by Step

Since for any $a\gt0$, $|x|=a$ implies $x=a$ OR $x=-a$, then the given equation, $ |-3x+8|=1 ,$ is equivalent to \begin{array}{l}\require{cancel} -3x+8=1 \text{ OR } -3x+8=-1 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} -3x+8=1 \\\\ -3x=1-8 \\\\ -3x=-7 \\\\ x=\dfrac{-7}{-3} \\\\ x=\dfrac{7}{3} \\\\\text{ OR }\\\\ -3x+8=-1 \\\\ -3x=-1-8 \\\\ -3x=-9 \\\\ x=\dfrac{-9}{-3} \\\\ x=3 .\end{array} Hence, the solutions are $ x=\left\{ \dfrac{7}{3},3 \right\} .$
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